# Tutor Hunt Questions

“A 0.050 mol sample of a hydrocarbon was burned in excess oxygen.

The products were 3.60 g of water and 6.60 g of carbon dioxide.

Calculate the number of moles of carbon dioxide produced.

Relative atomic masses: C = 12 O = 16.

When the hydrocarbon was burned 0.20 mol of water were produced.

How many moles of hydrogen atoms are there in 0.20 mol of water?

The products were 3.60 g of water and 6.60 g of carbon dioxide.

Calculate the number of moles of carbon dioxide produced.

Relative atomic masses: C = 12 O = 16.

When the hydrocarbon was burned 0.20 mol of water were produced.

How many moles of hydrogen atoms are there in 0.20 mol of water?

15 months ago

Chemistry Question asked by Carley

6.6g of CO2 produced during the combustion, therefore
RMM CO2 = 44
mol of CO2 produced = 6.6g/44g/mol = 0.15 mol
Water formula = H20
0.2 mol of water produced and the RAM for H=1, RMM, H2O =18
so, 2/18=H/0.2 H=0.0222 mol

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## 2 Answers

CH4 + 2O² ==> 2H20 + CO2

(0.05m) (3.60g) (6.60g)

[hydrocarbon + excess oxygen]

No of moles of CO2 produced == mass/molarmass =6.60g/44g/mol

=0.15mole

[CO2(molarmass)= 12+16(2)=44g/mol]

(ii) moles of hydrogen in 0.2mole of H20

1mole of H2O contains 2moles of H

0.20mole of H2O will contain

0.2mole * 2moles = 0.4mole of Hydrogen

(0.05m) (3.60g) (6.60g)

[hydrocarbon + excess oxygen]

No of moles of CO2 produced == mass/molarmass =6.60g/44g/mol

=0.15mole

[CO2(molarmass)= 12+16(2)=44g/mol]

(ii) moles of hydrogen in 0.2mole of H20

1mole of H2O contains 2moles of H

0.20mole of H2O will contain

0.2mole * 2moles = 0.4mole of Hydrogen

Answered by Zainab | 15 months ago

We known that there is 0.05 mol of an unknown hydrocarbon we`ll call `X` and that it`s combustion forms 3.6 g of water and 6.6 g of carbon dioxide.

It would be sensible to convert these amounts to moles using:

amount (moles) = mass (grams) / relative atomic mass (g mol^-1)

therefore for H2O:

moles = 3.6 / 18 = 0.2

similarly for CO2:

moles = 6.6 / 44 = 0.15

now we can write the equation as:

0.05 X + x` O2 -> 0.15 CO2 + 0.2 H2O

this can be simplified by dividing through by 0.05

(this is probably the step that will be most useful to you)

giving:

1 X + x O2 -> 3 CO2 + 4 H2O

(try finishing from here before reading the rest)

____________________

All the carbon in the carbon dioxide and all of the hydrogen in the water must have come from the hydrocarbon.

This means 3 x 1 = 3 moles of C and 4 x 2 = 8 moles of H.

In a single molecule of X there are 3/1 = 3 Cs and 8/1 = 8 Hs

(Dividing by 1 seems pointless but it shows the procedure to follow if there were more than one mol of X in the initial equation)

This clearly gives C3H8 or propane as hydrocarbon X

It would be sensible to convert these amounts to moles using:

amount (moles) = mass (grams) / relative atomic mass (g mol^-1)

therefore for H2O:

moles = 3.6 / 18 = 0.2

similarly for CO2:

moles = 6.6 / 44 = 0.15

now we can write the equation as:

0.05 X + x` O2 -> 0.15 CO2 + 0.2 H2O

this can be simplified by dividing through by 0.05

(this is probably the step that will be most useful to you)

giving:

1 X + x O2 -> 3 CO2 + 4 H2O

(try finishing from here before reading the rest)

____________________

All the carbon in the carbon dioxide and all of the hydrogen in the water must have come from the hydrocarbon.

This means 3 x 1 = 3 moles of C and 4 x 2 = 8 moles of H.

In a single molecule of X there are 3/1 = 3 Cs and 8/1 = 8 Hs

(Dividing by 1 seems pointless but it shows the procedure to follow if there were more than one mol of X in the initial equation)

This clearly gives C3H8 or propane as hydrocarbon X

Answered by Gregory | 15 months ago